Answer :
The school's claim group of answer options has a p-value for a test of 0.1539.
The notation and data are provided.
The sample is chosen at random: n = 130
The estimated proportion of students who intend to pursue general practice is: [tex]\bar p = 32\% = \frac{28}{10} = 0.32[/tex]
The value to be tested is: p₀ = 28% = 28/100 = 0.28
The statistic (variable of interest): z
The p-value (variable of interest): [tex]p_{v}[/tex]
Concepts and formulas used
In order to verify the assertion that the true proportion is greater than 0.28, we must test a hypothesis:
p ≤ 0.28 indicates a null hypothesis.
p > 0.28 is an alternative hypothesis.
The z statistic is required when doing a proportion test because it is provided by:
[tex]z= \frac{\bar p \;-\; p_{0}}{\sqrt{\frac{p_{0}\; (1\; - \; p_{0} ) }{n} } } \;\;\;\;\;\; ................ep.\; 1[/tex]
Determine the statistic.
Since we have all the necessary information, we can change equation 1 to read as follows:
[tex]z= \frac{0.32 \;-\; 0.28}{\sqrt{\frac{0.28\; (1\; - \; 0.28) }{130} } } \;\;\;\;\;\; \\\\z = \frac{0.04}{\sqrt{\frac{0.28\; * \; 0.72 }{130} } } \\\\z = \frac{0.04}{\sqrt{\frac{0.2016 }{130} } }\\\\z = 1.01575 \;or\; 1.02[/tex]
Statistical decision
The p-value for this test would then be calculated as the next step.
The p-value for this right-tailed test would be:
[tex]p_{v} = P \;(z > 1.02) = 0.1539[/tex]
Since we are determining whether the mean is greater than a value, with z = 1.01575, the p-value is obtained using a z-distribution calculator with a right-tailed test, and it is 0.1539.
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